1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Evaluate! surface integral over surface

  1. May 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Evaluate the surface integral of G over the surface S
    S is the parabolic cylinder y=2x^2, 0=< x =<5, 0=< z =<5
    G(x,y,z)=6x

    Answer is one of the following:
    1. (15/8)*(401sqrt(401)-1)
    2. (5/8)*(401sqrt(401)-1)
    3. (15/8)*(401sqrt(401)+1)
    4. (5/8)*(401sqrt(401)+1)

    2. Relevant equations


    3. The attempt at a solution
    Let f(x,y,z)=y-2x^2=0
    n=gradf/||gradf||=(-4xi+1j+0k)/sqrt(16x^2+1)
    n*=j
    G.n=-24x^2/sqrt(16x^2+1)
    n.n*=1/sqrt(16x^2+1)
    therefore the double integral over S = SS (G.n/n.n*) dzdx
    solving the double integral gets -5000
     
  2. jcsd
  3. May 29, 2012 #2
    Let me try ot reason through what you have. If at the end you were integrating dzdx, I'll assume I should try to paramet(e)rize the surface in x and z.

    Then r(x,z)=(x,2x^2,z). Then r_x=(1,4x,0), while r_z=(0,0,1).

    Then cross product is (r_x)x(r_z)=(4x,-1,0).

    Then the infinitesimal area on the surface is given by

    dS=sqrt(16x^2+1)dxdz.

    Now we want to integrate the SCALAR G(x,y,z)=4x against the area.

    That is, integrate G dS.

    So [itex]\int_{x=0}^5\int_{z=0}^54x\sqrt{16x^2+1}\ dzdx.[/itex]

    So my first guess is, you are mixing up VECTOR integrals with SCALAR integrals. You might compare them and their derivations.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook